Conditions (3) and (5) combined with Lemma REF  and Theorem REF  provide that \(\mathbf {K}\)  is the union of three disjoint nonempty \(\Phi \)  and \(\Phi ^{-1}\) -invariant subsets \(\mathbf {K}_1, \mathbf {K}_2\) , and \(\mathbf {O}\)  with the properties

The sets \(\mathbf {K}_1\)  and \(\mathbf {K}_2\)  are closed in \(\mathbf {K}\) .

\(\mathbf {T}\)  is invertible on \(C(\mathbf {K}_2)\)  and \(\rho (\mathbf {T}^{-1}, C(\mathbf {K}_2)) < 1\) .

\(\rho (\mathbf {T}, C(\mathbf {K}_1)) < 1\) .

If \(\mathbf {V}_1\)  and \(\mathbf {V}_2\)  are open neighborhoods of \(\mathbf {K}_1\)  and \(\mathbf {K}_2\) , respectively, and \(\mathbf {E}\)  is a closed subset of \(\mathbf {O}\)  than there is an \(m \in {N}\)  such that for any \(n \ge m\)  we have
\(\Phi ^n(\mathbf {E}) \subseteq \mathbf {V}_2 \; \text{and} \; \Phi ^{-n}(\mathbf {E}) \subseteq \mathbf {V}_1.\) 

Let \(p: \mathbf {K} \rightarrow K\)  be the map \(p(\mathbf {k})=k_0\)  introduced in the proof of Lemma REF  and let
\(K_1 = p(\mathbf {K_1})\) , \(K_2 = p(\mathbf {K}_2)\) , and \(O = p(\mathbf {O})\) . We proceed with proving the statements (a) - (i) of the lemma.
(a). Obviously the set \(K_1\)  is closed. It follows from the equalities \(\Phi (\mathbf {K}_1) = \mathbf {K}_1\) , \(\Phi (\mathbf {O}) = \mathbf {O}\) , and \(\Phi (\mathbf {K}_2) = \mathbf {K}_2\)  and the definition of the map \(p\)  that \(\varphi (K_1) = K_1\) , \(\varphi (O) = O\) , and \(\varphi (K_2) = K_2\) . Moreover, it follows from properties (II) - (IV) that \(K_1\)  does not intersect with \(K_2 \cup O\)  and therefore, \(\varphi (K_1) = K_1 = \varphi ^{-1}(K_1)\) .
(b). Follows from (III) and the definition of \(p\) .
(c). The inequality \(\rho (\mathbf {T}^{-1}, C(\mathbf {K}_2)) < 1\)  follows from (II). It follows from the fact that the operator \(\lambda I - T\)  is lower semi-Fredholm and Theorem REF  that \(\varphi \)  is an almost homeomorphism of \(K_2\) .
(d). It follows from Baire Category Theorem and the fact that \(K_1 \cap F = \emptyset \)  that there is an \(n \in {N}\)  such that \(Int \, \varphi ^{(-n)}(K_2) \ne \emptyset \) . Then, because \(\varphi \)  is open, \(Int \, K_2 \ne \emptyset \) . Assume, contrary to our claim, that \(Int \, K_2 K_2\) . Let \(\tilde{K} = K \setminus \bigcup \limits _{n=0}^\infty \varphi ^{(-n)}(Int \, K_2)\) . Notice that \(\tilde{K} = K_1 \cup \bigcup \limits _{n=0}^\infty \varphi ^{(-n)}(E)\)  where \(E = K_2 \setminus Int \, K_2\) . Applying again Baire category theorem we see that there is an \(n \in {N}\)  such that \(Int_{\tilde{K}} \varphi ^{(-n)}(E) \ne \emptyset \) . Therefore, there is an open subset \(V\)  of \(K\)  such that \(V \cap \varphi ^{(-n)} \subset \varphi ^{(-n)}\) . Then \(\varphi ^n(V) \cap E \subset E \ne \emptyset \) . The set \(Int \, K_2 \cup \varphi ^n(V) \cap E\)  is open in \(K\)  and \( Int \, K_2 \subsetneqq Int \, K_2 \cup (\varphi ^n(V) \cap E)\) , a contradiction.
(e) - (h). For brevity let \(F = \bigcup \limits _{n=0}^\infty \varphi ^{(-n)}(K_2)\) . We assume that \(Q \ne \emptyset \) , otherwise the statement is trivial.
Let \(k \in Q\)  and let \(E = p^{-1}(\lbrace k\rbrace )\) . \(E\)  is a compact subset of \(\mathbf {O}\)  and therefore by Theorem REF  for arbitrary open neighborhoods \(\mathbf {V}_1\)  and \(\mathbf {V}_2\)  of, respectively, \(\mathbf {K}_1\)  and \(\mathbf {K}_2\)  there is an \(N \in {N}\)  such that \(\Phi ^n(E) \subseteq \mathbf {V}_2\)  and \(\Phi ^{-n}(E) \subseteq \mathbf {V}_1\)  for any \(n \ge N\) . Define a function \(f\)  on \(K\)  as follows,
\(f(x) ={\left\lbrace \begin{array}{ll} 1 , & \text{if $x = k$,}\\\lambda ^i/w_i(k) , & \text{if $x = \varphi ^{(i)}(k), i \in {N}$,}\\\lambda ^{-i} w_i(x) , & \text{if $x \in \varphi ^{(-i)}(\lbrace k\rbrace ), i \in {N}$,}\\0, &\text{ otherwise,}\end{array}\right.}\) 
It is immediate to see that \(f\)  is an element of \(C(K)^{\prime \prime }\)  and \(T^{\prime \prime } f = \lambda f\) . Because we assumed that \(\lambda I - T\)  is a semi-Fredholm operator it follows that (see e.g. [1]}) \(\ker (\lambda I - T) \ne \mathbf {0}\) . Let \(f \in C(K)\) , \(f \ne 0\) , and \(Tf = \lambda f\) . Let \(U = \lbrace k \in K : \; f(k) \ne 0\rbrace \) . It follows from \(Tf = \lambda f\) , \(\lambda \ne 0\) , and \(w \in C(K)^{-1}\)  that
\(\varphi (U) = U = \varphi ^{(-1)}(U) \; \text{and} \; U \cap (K_1 \cup F) = \emptyset .\) 
By Zorn's lemma there is a maximal by inclusion open subset \(V\)  of \(K\)  with properties (REF ). We claim that
\(V \cup K_1 \cup F = K.\) 
To prove it consider the compact space \(G = K \setminus V\) . It follows from (REF ) that \(\varphi (G) = G\)  and therefore the operator \(T\)  is defined on \(C(G)\) . Moreover, the operator \(\lambda I - T\)  is lower semi-Fredholm. Assume that \(K_1 \cup F G\) . Then our previous reasoning shows that there is \(g \in C(G)\)  such that \(g \ne 0\)  and \(Tg = \lambda g\) . Let \(H = \lbrace k \in G : g(k) \ne 0\rbrace \) . Then the set \(V \cup H\)  is open in \(K\)  and satisfies (REF ) in contradiction with the maximality of \(V\) .
We claim that for any compact subset \(W\)  of \(V\)  and for any open neighborhoods \(U_1\)  and \(U_2\)  of \(K_1\)  and \(K_2\) , respectively, there is an \(N \in {N}\)  such that for any \(n > N\)  we have
\(\varphi ^n(W) \subset U_2 \; \text{and} \; \varphi ^{(-n)}(W) \subset U_1.\) 
Indeed, (REF ) follows from the fact that \(p^{(-1)}(W)\)  is a compact subset of \(\mathbf {O}\) .
Notice that it follows from (REF ) that \(cl V \cap K_2 \ne \emptyset \) .
Next we claim that there is an open neighborhood \(Q\)  of \(K_2\)  in \(K\)  such that \(\varphi (Q \cap V) \subseteq Q \cap V\)  and the restriction of \(\varphi \)  on \(Q \cap V\)  is a homeomorphism of \(Q \cap V\)  onto its image. First notice that \(\ker {(\lambda I - T^{\prime })} = \ker {((\lambda I - T^{\prime }), C(K_2))} \) . Indeed, let \(\mu \in C(K)^{\prime }\)  be such that \(T"\mu = \mu \) . Let \(U\)  be an open neighborhood of \(K_2\)  and let \(f \in C(K)\)  be such that \(supp \, f \subseteq K \setminus U\) . Then \(\int f d\mu = \lambda ^{-n} \int T^n fd\mu \mathop \rightarrow \limits _{n \rightarrow \infty } 0\) , because \(supp \, T^n f \subseteq K \setminus \varphi ^{(-n)}(U)\) , \(\bigcap \limits _{n=1}^\infty (K \setminus \varphi ^{(-n)}(U)) = K_1\) , and \(\rho (T, C(K_1)) < |\lambda |\) . It follows that \(supp \, \mu \subseteq K_2\) .
Assume, contrary to our claim, that for any open neighborhood \(Q\)  of \(K_2\)  there are \(p, q \in Q \cap V\)  such that \(\varphi (p) = \varphi (q)\) . The proof of Lemma 5.9 in [2]} shows that there is a sequence \(\mu _n \in (C(K)^{\prime }\)  such that \(\Vert \mu _n\Vert  = 1\) , \(T^{\prime }\mu _n - \lambda \mu _n \rightarrow 0\) , and for each \(n\)  the measure \(\mu _n\)  is a finite linear combination of point measures \(\delta _{s_i}\)  where \(S_i \in V\) . Because the operator \(\lambda I - T^{\prime }\)  is lower semi-Fredholm the sequence \(\mu _n\)  must contain a norm convergent subsequence. Let the limit of such a subsequence be \(\nu \)  then \(T^{\prime }\nu = \nu \)  and by our previous step \(supp \, \nu \subseteq K_2\) . That means that for every \(n\)  we have \(|\nu | \wedge |\mu _n| =0\) , a contradiction.
Thus, there is an open neighborhood \(Q\)  of \(K_2\)  such that the restriction of \(\varphi \)  on \(Q \cap V\)  is one-to-one. The proofs of Lemma 5.10 and Corollary 5.11 in [2]} show that we can choose \(Q\)  in such a way that \(\varphi (V \cap Q) \subset V \cap Q\) .
